Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
727 views
in Technique[技术] by (71.8m points)

python - Linear Regression from Time Series Pandas

I would like to get a regression with a time series as a predictor and I'm trying to follow the answer give on this SO answer (OLS with pandas: datetime index as predictor) but it no longer seems to work to the best of my knowledge.

Am I missing something or is there a new way to do this?

import pandas as pd

rng = pd.date_range('1/1/2011', periods=4, freq='H')       
s = pd.Series(range(4), index = rng)                                                                      
z = s.reset_index()

pd.ols(x=z["index"], y=z[0]) 

I'm getting this error. The error is explanatory but I'm wondering what I'm missing in reimplementing a solution that worked before.

TypeError: cannot astype a datetimelike from [datetime64[ns]] to [float64]

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I'm not sure why pd.ols is so picky there (it does appear to me that you followed the example correctly). I suspect this is due to changes in how pandas handles or stores datetime indexes but am too lazy to explore this further. Anyway, since your datetime variable differs only in the hour, you could just extract the hour with a dt accessor:

pd.ols(x=pd.to_datetime(z["index"]).dt.hour, y=z[0])

However, that gives you an r-squared of 1, since your model is overspecified with the inclusion of an intercept (and y being a linear function of x). You could change the range to np.random.randn and then you'd get something that looks like normal regression results.

In [6]: z = pd.Series(np.random.randn(4), index = rng).reset_index()                                                               
        pd.ols(x=pd.to_datetime(z["index"]).dt.hour, y=z[0])
Out[6]: 

-------------------------Summary of Regression Analysis-------------------------

Formula: Y ~ <x> + <intercept>

Number of Observations:         4
Number of Degrees of Freedom:   2

R-squared:         0.7743
Adj R-squared:     0.6615

Rmse:              0.5156

F-stat (1, 2):     6.8626, p-value:     0.1200

Degrees of Freedom: model 1, resid 2

-----------------------Summary of Estimated Coefficients------------------------
      Variable       Coef    Std Err     t-stat    p-value    CI 2.5%   CI 97.5%
--------------------------------------------------------------------------------
             x    -0.6040     0.2306      -2.62     0.1200    -1.0560    -0.1521
     intercept     0.2915     0.4314       0.68     0.5689    -0.5540     1.1370
---------------------------------End of Summary---------------------------------

Alternatively, you could convert the index to an integer, although I found this didn't work very well (I'm assuming because the integers represent nanoseconds since the epoch or something like that, and hence are very large and cause precision issues), but converting to integer and dividing by a trillion or so did work and gave essentially the same results as using dt.hour (i.e. same r-squared):

pd.ols(x=pd.to_datetime(z["index"]).astype(int)/1e12, y=z[0])

Source of the error message

FWIW, it looks like that error message is coming from something like this:

pd.to_datetime(z["index"]).astype(float)

Although a fairly obvious workaround is this:

pd.to_datetime(z["index"]).astype(int).astype(float)

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...